Let [t] be the greatest integer less than or | vedclass

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(C) We know that $\lim_{x \rightarrow 0^{+}} x[\frac{k}{x}] = k$ for any constant $k > 0$. Applying this to the given limit: $\lim_{x \rightarrow 0^{+

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) We know that $\lim_{x ightarrow 0^{+}} x[\frac{k}{x}] = k$ for any constant $k > 0$. Applying this to the given limit: $\lim_{x ightarrow 0^{+}} x \sum_{k=1}^{p} [\frac{k}{x}] = \sum_{k=1}^{p} k = \frac{p(p+1)}{2}$. Similarly,$\lim_{x ightarrow 0^{+}} x^2 [\frac{k^2}{x^2}] = k^2$. So,$\lim_{x}$ ${ightarrow 0^{+}} x^2 \sum_{k=1}^{9} [\frac{k^2}{x^2}] = \sum_{k=1}^{9} k^2 = \frac{9(9+1)(2 	imes 9 + 1)}{6} = \frac{9 	imes 10 	imes 19}{6} = 285$. The inequality becomes: $\frac{p(p+1)}{2} - 285 \geq 1$ $\frac{p(p+1)}{2} \geq 286$ $p(p+1) \geq 572$. For $p=23$,$23 	imes 24 = 552 For $p=24$,$24 	imes 25 = 600 \geq 572$. Thus,the least natural value of $p$ is $24$.

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